The Biclique Covering Number of Grids

The biclique covering number of grids

Krystal Guo1 Tony Huynh2 Marco Macchia3 Département de Mathématique, Universitélibre de Bruxelles, Belgium. (1) [email protected] (2) [email protected] (3) [email protected]

Submitted: Nov 16, 2018; Accepted: Oct 28, 2019; Published: Nov 8, 2019 c The authors. Released under the CC BY-ND license (International 4.0).

Abstract We determine the exact value of the biclique covering number for all grid graphs. Mathematics Subject Classiﬁcations: 05C70

1 Introduction

Let G be a graph. A biclique of G is a complete bipartite subgraph. The biclique covering number of G, denoted bc(G), is the minimum number of bicliques of G required to cover the edges of G. The biclique covering number is studied in ﬁelds as diverse as polyhedral combinatorics [10,2], biology [11], and communication complexity [12], where it is also known as bipartite dimension or rectangle covering number. Computing the biclique covering number is a classic NP-hard problem. Indeed, decid- ing if bc(G) 6 k appears as problem GT18 in Garey and Johnson [7]. It is also NP-hard to approximate within a factor of n1−ε, where n is the number of vertices of G (see [4]). As such, there are very few classes of graphs for which we know the biclique covering number exactly. For example, it is well-known that the biclique covering number of the complete graph Kn is dlog2 ne (see [5] for a proof). Note that the minimum number of bicliques that partition E(Kn) is n−1 by the Graham-Pollak theorem [9]. This result was later extended to biclique coverings C of Kn such that each edge is in at most k bicliques of C. Alon [1] showed that the minimum number of bicliques in such coverings is Θ(kn1/k). Two more classes of graphs for which we know the biclique covering number exactly − − are K2n and Kn,n, which are the graphs obtained from K2n and Kn,n by deleting the edges − of a perfect matching. The biclique covering number of K2n is dlog2 ne (see [13]) and the biclique covering number of K− is the smallest k for which n k (see [6,3]). n,n 6 bk/2c

the electronic journal of combinatorics 26(4) (2019), #P4.27 1 In this paper, we determine the biclique covering number for all grids. Let Gp,q be the 0 0 p × q grid. Recall that Gp,q has vertex set [q] × [p], where (a, b) is adjacent to (a , b ) if and only if |a − a0| + |b − b0| = 1. The following is our main result. Theorem 1. For all integers 1 6 p 6 q,  pq  2 − 1, if p is even and q − 1 = k(p − 1) + 2` for some integers 0 6 ` < k; bc(Gp,q) = pq  2 , otherwise.

Since Gp,q ' Gq,p, our main result determines the biclique covering number of all grid graphs. This settles an open problem raised by Denis Cornaz at the 9th Carg`eseWorkshop on Combinatorial Optimization in 2018. Note that for p even and q > p(p−1), Theorem1 pq implies that bc(Gp,q) = /2 − 1.

2 Main result

Since we are only interested in biclique covers of minimum size, we may assume that biclique covers only consist of maximal bicliques (under edge inclusion). For brevity, we call such biclique covers simply covers. For Gp,q, this implies that covers consist of elements which are isomorphic to K1,3,K1,4, or the 4-cycle. We ﬁrst establish the following upper bound. pq Lemma 2. For all integers 1 6 p 6 q, bc(Gp,q) 6 b /2c.

pq Proof. Observe that Gp,q is a bipartite graph with bipartition (X,Y ) where |X| = b /2c and |Y | = dpq/2e. Therefore taking the set of stars centered at vertices in X gives a cover of size bpq/2c. See Figure1.

Figure 1: The stars centered at black vertices are a cover.

We now establish the following lower bounds. Lemma 3. For all integers 1 6 p 6 q, pq /2 6 bc(Gp,q), if p is odd. pq /2 − 1 6 bc(Gp,q), if p is even.

the electronic journal of combinatorics 26(4) (2019), #P4.27 2 Proof. Let 1 6 p 6 q. We deﬁne a special subset S(Gp,q) of edges of Gp,q inductively as follows. If p ∈ {1, 2}, we let S(Gp,q) be the set of all horizontal edges of Gp,q. If p > 3, let Out be the outer cycle of Gp,q and deﬁne S(Gp,q) = E(Out) ∪ S(Gp,q − V (Out)). See Figure 2a and Figure 2b. An easy induction gives

(i) |S(Gp,q)| = pq − 1, if p is odd.

(ii) |S(Gp,q)| = pq − 2, if p is even.

(a) S(Gp,q) for p even. (b) S(Gp,q) for p odd. Figure 2

On the other hand, every biclique of Gp,q contains at most 2 edges of S(Gp,q). Therefore, |S(G )| bc(Gp,q) > d p,q /2e, which completes the proof.

pq Next, we give values of p and q for which bc(Gp,q) = b /2c − 1. pq Lemma 4. bc(Gp,q) = /2 − 1 if p is even and q − 1 = k(p − 1) + 2` for some integers 0 6 ` < k.

Proof. By Lemma3, it suﬃces to construct a cover C of size pq/2 − 1. Since q − 1 = k(p − 1) + 2` for some integers 0 6 ` < k, we can decompose Gp,q as k copies of Gp,p and ` copies of Gp,3 such that each of the ` copies of Gp,3 is between two copies of Gp,p. Since p2 p is even, Gp,p has two covers of size /2 − 1, whose set of 4-cycles are the 4-cycles of the two diagonals of Gp,p, respectively. See Figure3 for the case p = 6. The cover C is constructed as follows. For every two consecutive copies of Gp,p, we use one cover from Figure3 on one copy and the other cover from Figure3 on the other copy. p Note that this results in /2 − 1 bicliques that are in both copies. For every copy of Gp,3 p2 between two copies of Gp,p we again use the two covers of size /2 − 1 on the two copies of p Gp,p, and then we use /2 + 1 bicliques in Gp,3 to cover the remaining edges, see Figure4. The total construction when p = 6 and q = 25 is given in Figure5. Observe that: p2 p p p pq |C| = k − 1 + ` + 1 − (k − ` − 1) − 1 = k(p − 1) + 2` + 1 − 1 = − 1 . 2 2 2 2 2 We now establish that we may choose a cover with special properties, before completing the proof of the main theorem. Recall that Out is the outer cycle of Gp,q. Let C be a cover of Gp,q.A boundary element of C is an element of C containing at least one edge of

the electronic journal of combinatorics 26(4) (2019), #P4.27 3 (a) (b)

Figure 3: Two covers of G6,6 of size 17.

G8,8 G8,3 G8,8

Figure 4: Cover of G8,17 of size 67.

G6,6 G6,3 G6,6 G6,3 G6,6 G6,6

Figure 5: Cover of G6,25 of size 74.

Out.A boundary 4-cycle is a boundary element that is a 4-cycle and a boundary star is a boundary element that is a star.

Lemma 5. For every cover of Gp,q, there exists a cover C of the same size with the following

the electronic journal of combinatorics 26(4) (2019), #P4.27 4 properties: (i) all boundary stars of C are pairwise edge-disjoint; (ii) no edge is contained in both a boundary 4-cycle and a boundary star of C. Proof. If two boundary stars meet in an edge, we replace them with a star and a 4-cycle which cover a superset of edges, as in Figure6. If a boundary 4-cycle and a boundary star are not edge-disjoint, then we replace them with two 4-cycles which cover a superset of edges, as in Figure7.

(a) (b)

Figure 6

(a) (b)

Figure 7

By repeatedly performing these two replacement rules, we eventually obtain a cover C satisfying (i) and (ii). We now complete the proof of Theorem1 by establishing the following converse to Lemma4. pq Lemma 6. Let 1 6 p 6 q, with p even. If Gp,q has a cover of size /2 − 1, then q − 1 = k(p − 1) + 2` for some integers 0 6 ` < k. Proof. The lemma clearly holds if p = 2, so we may assume p > 4. Let C be a cover of pq Gp,q of size /2 − 1 satisfying properties (i) and (ii) of Lemma5. We begin by defining some objects required for the proof. Let Out be the outer cycle of Gp,q. The corners of Gp,q are the vertices (1, 1), (1, p), (q, 1), and (q, p). Let H be the subgraph of Gp,q induced by the edges of the boundary 4-cycles of C.A fence is a connected component of H. The size of a fence is the number of boundary 4-cycles it contains. A link is a connected component of Out \ E(H) containing at least two vertices and no corners. See Figure8 for an illustration of fences and links. Suppose L is a link contained on the topmost path of Gp,q. Let Lleft be the path starting from the left end of L by first proceeding down twice and then alternating between the electronic journal of combinatorics 26(4) (2019), #P4.27 5 proceeding right and down. We define Lright by interchanging left and right in the definition of Lleft. The staircase generated by L is the subgraph of Gp,q contained in the region bounded by L, Lleft, Lright, and (possibly) a subpath of the bottommost path of Gp,q, see Figure 9a and Figure 9b. Staircases for links contained in the leftmost, rightmost, and bottommost paths of Gp,q are defined by rotating Gp,q so that the link is on the topmost path, applying the above definition, and then rotating back. The length of a path is its number of edges, and the length of a staircase is the length of the link that generates it. An edge is thick if it is covered by at least two bicliques in C. A pyramid P is a staircase of length 2p − 4 containing a size-2 fence B such that the thick edge in B is the only thick edge in P, see Figure 10. We call B the tip of the pyramid.

Figure 8: Fences are shaded. Links are shown in blue.

(a) The staircase generated by a link of length 10 is shown in (b) The staircase generated blue. In this case, the staircase includes part of the bottom- by a link of length 4 is shown most path of the grid. in blue.

Figure 9

Claim 7. Every staircase contains at least one thick edge. Proof. Let L be a link of length k connecting two fences B and B0 (see Figure 11a) and let S be the staircase generated by L. Towards a contradiction, suppose S does not contain a thick edge. Since C satisﬁes properties (i) and (ii) of Lemma5, k is even and P is covered by exactly k/2 boundary stars as in Figure 11b. Let E1 be the set of edges in S with one end on L and not covered by a boundary element of C (blue edges in Figure 11b). Each edge e in E1 must be covered by a K1,4, else e is thick, see Figure 11c. Let e1 and e2 be the

the electronic journal of combinatorics 26(4) (2019), #P4.27 6 Figure 10: A pyramid and its tip (in dark blue). The red edge is the only thick edge.

two horizontal edges of S which intersect (V (B)∪V (B0))\V (L) (blue edges in Figure 11c). Since neither e1 nor e2 are thick, they must be covered by 4-cycles as in Figure 11d. By repeating this argument, we either ﬁnd a thick edge (red edge in Figure 11e), or we reach the bottommost path of Gp,q. Let F be the set of edges f ∈ E(S) such that f is covered by a K1,4 of C and f has exactly one endpoint on the bottommost path of Gp,q. By the above argument, F is non-empty. Moreover, each edge in F is thick (see Figure 11f).

B L B0

(a) (b) (c)

(d) (e) (f)

Figure 11

A double staircase is a pair (S, S0), where S and S0 are distinct staircases such that there is exactly one thick edge in S ∪ S0. Suppose S has length 2a and S0 has length 2b. Observe that either a + b = p − 2 or a + b = p and C must cover S ∪ S0 as in Figures 12a and 12b, respectively.

the electronic journal of combinatorics 26(4) (2019), #P4.27 7 (a) (b)

Figure 12: Examples of double staircases. The red edge is the only thick edge.

We deﬁne the waste of C to be

w(C) := 4|C| − |E(Gp,q)|. (1)

Let τ be the number of copies of K1,3 that appear in C and ti be the number of edges of Gp,q that are covered exactly i times by C. By double counting, we obtain:

w(C) = τ + t2 + 2t3 + 3t4. (2) pq Since |C| = /2 − 1 and |E(Gp,q)| = 2pq − p − q,(1) immediately gives the following: w(C) = p + q − 4 . (3)

Note that a fence can have size at most N := 2p + 2q − 8. For i ∈ [N], let bi be the number of fences of size i of C.

Claim 8. bN = 0.

Proof. If bN = 1, then C contains 2p + 2q − 8 boundary 4-cycles. These boundary 4-cycles cover 6p + 6q − 24 edges of Gp,q. Thus, 2pq − 7p − 7q + 24 edges are not covered by boundary 4-cycles. It follows that 2pq − 7p − 7q + 24 pq p q pq |C| 2p + 2q − 8 + = + + − 2 , > 4 2 4 4 > 2 where the last inequality follows since 4 6 p 6 q. However, this contradicts |C| = pq/2 − 1. Let β be the number of edges that are covered twice by boundary 4-cycles of C, and let c ∈ [4] be the number of corners of Gp,q covered by the set of fences. We now provide a lower bound on the waste of C. N−1 c b1 X i Claim 9. w(C) β + τ = p + q − 2 − − + − 1 b . > 2 2 2 i i=3 Proof. The inequality follows immediately from (2). For the equality, ﬁrst observe β = PN−1 i=1 (i − 1)bi. Next, because each boundary star covers two edges of Out, each edge of Out not covered by a fence contributes 1/2 to τ. Since Out contains 2p + 2q − 4 edges and

the electronic journal of combinatorics 26(4) (2019), #P4.27 8 PN−1 the set of fences cover c + i=1 ibi edges of Out, we have N−1 N−1 X 1 X β + τ = (i − 1)b + 2p + 2q − 4 − c − ib i 2 i i=1 i=1 N−1 c b1 X i = p + q − 2 − − + − 1 b , 2 2 2 i i=3 as desired. 0 0 Claim 10. The set of staircases of C can be enumerated as P1,..., Pn, S1, S1,..., Sm, Sm 0 such that Pi is a pyramid for all i ∈ [n] and (Sj, Sj) is a double staircase for all j ∈ [m]. Moreover, b2 = n, bi = 0 for all i > 3, and w(C) = τ + n + m. PN−1 Proof. The number of staircases of C is at least i=1 bi + c − 4. At most b2 fences can be PN−1 the tips of pyramids, so at least b1 + i=3 bi + c − 4 staircases are not pyramids. Each of these staircases contains a thick edge by Claim7, and each thick edge can be in at most PN−1 2 staircases. Therefore, these staircases contribute at least (b1 + i=3 bi + c − 4)/2 to the waste of C which is not counted in β + τ. By Claim9, N−1 N−1 c b1 X i 1 X w(C) p + q − 2 − − + − 1 b + b + b + c − 4 > 2 2 2 i 2 1 i i=3 i=3 N−1 X i − 1 = p + q − 4 + b . 2 i i=3 By (3), w(C) = p+q−4. Hence, we must have bi = 0 for i > 3 and equality throughout the above argument. In particular, there are exactly b2 pyramids. The remaining staircases each contain exactly one thick edge, and each of these thick edges is in exactly two staircases. This gives the required enumeration of the staircases of C and implies w(C) = τ + n + m. Claim 11. q − 1 = k(p − 1) + 2` for some integers 0 6 ` < k.

Proof. Let L be the leftmost path of Gp,q. Since q > p, no link contained in L can generate a pyramid or be part of a double staircase. By Claim 10, no link is contained in L. It follows that L intersects at most one fence. Since p is even, it follows that L intersects exactly one fence Bleft. Since all fences are of size 1 or 2 and p is even, Bleft is of size 1 and L \ E(Bleft) is the disjoint union of two even-length paths Pa and Pb. We assume that Pa is above Pb and that Pa and Pb have lengths a and b, respectively. By the same argument, the rightmost path of Gp,q intersects exactly one fence Bright. We ﬁrst consider the case when Bleft is a fence that does not contain a corner. Suppose L1 is the ﬁrst link along the topmost path of Gp,q and S1 is the staircase generated by L1. By Claim 10, there are no thick edges outside of staircases. Therefore, Bleft and the boundary stars force the leftmost portion of Gp,q to be covered by C as in Figure 13a.

the electronic journal of combinatorics 26(4) (2019), #P4.27 9 Bleft

Bright

(a) (b) Figure 13

0 0 This forces S1 to be in a double staircase (S1, S1), where S1 has length 2b and S1 has 0 length 2a (see Figures 14a and 14b), or S1 has length 2b + 2 and S1 has length 2a + 2 (see Figures 14c and 14d).

(a) (b)

Suppose L2 is the next link along the topmost path of Gp,q. Because of the previous 0 0 double staircase (S1, S1), S2 must be in a double staircase (S2, S2), where S2 has length 0 0 2a and S2 has length 2b (see Figures 14a and 14c), or S2 has length 2a + 2 and S2 has length 2b+2 (see Figures 14b and 14d). Repeating the argument, we obtain a sequence of 0 0 0 double staircases (S1, S1),..., (Sk−1, Sk−1), where (Sk−1, Sk−1) is the last double staircase. Note that it is possible that this sequence is empty, corresponding to the case that there are no staircases and k − 1 = 0. Again, because there are no thick edges outside of staircases, the fence Bright and the boundary stars force the rightmost portion of Gp,q to be covered as in Figure 13b. For all 0 0 i ∈ [k − 1], let ì and ì be the lengths of Si and Si respectively. Let ` be the number of 0 times that {ì, ì} = {2a + 2, 2b + 2}. It follows that q − 1 = k(p − 1) + 2`, as required. The case when Bleft contains a corner is treated exactly as above, resulting in the construction from Figure5. This completes the entire proof. the electronic journal of combinatorics 26(4) (2019), #P4.27 10 Acknowledgements We thank Denis Cornaz for suggesting this problem and for carefully reading an early draft of this paper. We also thank Matthias Walter for help in computing bc(Gp,q) for small values of p and q via SCIP [8]. This project is supported by ERC grant FOREFRONT (grant agreement no. 615640) funded by the European Research Council under the EU’s 7th Framework Programme (FP7/2007-2013).

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